SQL Create Unique Value Flag
There are lots of questions/answers about selecting unique values in a MySQL query but I haven't seen any on creating a unique value flag.
I'm have a customer_ID that can appear more than once in a query output. I want to create new column that flags whether the customer_ID is unique or not (0 or 1).
The output should look something like this:
ID | Customer ID | Unique_Flag
1 | 1234 | 1
2 | 2345 | 1
3 | 2345 | 0
4 | 5678 | 1
Please let me know if anybody needs clarifications.
mysql sql unique
edited Nov 23 '18 at 7:20
jarlh
28.4k52138
asked Nov 23 '18 at 4:45
Daniel Galletta
1591114
add a comment |
There are lots of questions/answers about selecting unique values in a MySQL query but I haven't seen any on creating a unique value flag.
I'm have a customer_ID that can appear more than once in a query output. I want to create new column that flags whether the customer_ID is unique or not (0 or 1).
The output should look something like this:
ID | Customer ID | Unique_Flag
1 | 1234 | 1
2 | 2345 | 1
3 | 2345 | 0
4 | 5678 | 1
Please let me know if anybody needs clarifications.
mysql sql unique
edited Nov 23 '18 at 7:20
jarlh
28.4k52138
asked Nov 23 '18 at 4:45
Daniel Galletta
1591114
add a comment |
There are lots of questions/answers about selecting unique values in a MySQL query but I haven't seen any on creating a unique value flag.
I'm have a customer_ID that can appear more than once in a query output. I want to create new column that flags whether the customer_ID is unique or not (0 or 1).
The output should look something like this:
ID | Customer ID | Unique_Flag
1 | 1234 | 1
2 | 2345 | 1
3 | 2345 | 0
4 | 5678 | 1
Please let me know if anybody needs clarifications.
mysql sql unique
edited Nov 23 '18 at 7:20
jarlh
28.4k52138
asked Nov 23 '18 at 4:45
Daniel Galletta
1591114
There are lots of questions/answers about selecting unique values in a MySQL query but I haven't seen any on creating a unique value flag.
I'm have a customer_ID that can appear more than once in a query output. I want to create new column that flags whether the customer_ID is unique or not (0 or 1).
The output should look something like this:
ID | Customer ID | Unique_Flag
1 | 1234 | 1
2 | 2345 | 1
3 | 2345 | 0
4 | 5678 | 1
Please let me know if anybody needs clarifications.
mysql sql unique
mysql sql unique
edited Nov 23 '18 at 7:20
jarlh
28.4k52138
asked Nov 23 '18 at 4:45
Daniel Galletta
1591114
edited Nov 23 '18 at 7:20
jarlh
28.4k52138
asked Nov 23 '18 at 4:45
Daniel Galletta
1591114
edited Nov 23 '18 at 7:20
jarlh
28.4k52138
edited Nov 23 '18 at 7:20
jarlh
28.4k52138
edited Nov 23 '18 at 7:20
jarlh
28.4k52138
28.4k52138
asked Nov 23 '18 at 4:45
Daniel Galletta
1591114
asked Nov 23 '18 at 4:45
Daniel Galletta
1591114
asked Nov 23 '18 at 4:45
Daniel Galletta
1591114
1591114
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
You seem to want to mark the first occurrence as unique, but not others. So, let's join in the comparison value:
select t.*,
(id = min_id) as is_first_occurrence
from t join
(select customer_id, min(id) as min_id
from t
group by customer_id
) tt
on t.customer_id = tt.customer_id;
For most people, a "unique" flag would mean that the overall count is "1", not that this is merely the first appearance. If that is what you want, then you can use similar logic:
select t.*,
(id = min_id) as is_first_occurrence,
(cnt = 1) as is_unique
from t join
(select customer_id, min(id) as min_id, count(*) as cnt
from t
group by customer_id
) tt
on t.customer_id = tt.customer_id;
And, in MySQL 8+, you would use window functions:
select t.*,
(row_number() over (partition by customer_id order by id) = 1) as is_first_occurrence,
(count(*) over (partition by customer_id) = 1) as is_unique
from t;
answered Nov 23 '18 at 12:39
Gordon Linoff
758k35291399
add a comment |
You can try below
select id,a.customerid, case when cnt=1 then 1 else 0 end as Unique_Flag
from tablename a
left join
(select customerid, count(*) as cnt from tablename
group by customerid
)b on a.customerid=b.customerid
answered Nov 23 '18 at 4:48
fa06
11k2917
add a comment |
You can use lead function as given below to get the required output.
SELECT ID, CUSTOMER_ID,
CASE
WHEN CUSTOMER_ID != CUSTOMER_ID_NEXT THEN 1
ELSE 0
END AS UNIQUE_FLAG FROM
(SELECT ID, CUSTOMER_ID,LEAD(CUSTOMER_ID, 1, 0) OVER (ORDER BY CUSTOMER_ID) AS CUSTOMER_ID_NEXT FROM TABLE)T
answered Nov 23 '18 at 9:36
Vivek Khandelwal
1246
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You seem to want to mark the first occurrence as unique, but not others. So, let's join in the comparison value:
select t.*,
(id = min_id) as is_first_occurrence
from t join
(select customer_id, min(id) as min_id
from t
group by customer_id
) tt
on t.customer_id = tt.customer_id;
For most people, a "unique" flag would mean that the overall count is "1", not that this is merely the first appearance. If that is what you want, then you can use similar logic:
select t.*,
(id = min_id) as is_first_occurrence,
(cnt = 1) as is_unique
from t join
(select customer_id, min(id) as min_id, count(*) as cnt
from t
group by customer_id
) tt
on t.customer_id = tt.customer_id;
And, in MySQL 8+, you would use window functions:
select t.*,
(row_number() over (partition by customer_id order by id) = 1) as is_first_occurrence,
(count(*) over (partition by customer_id) = 1) as is_unique
from t;
answered Nov 23 '18 at 12:39
Gordon Linoff
758k35291399
add a comment |
You seem to want to mark the first occurrence as unique, but not others. So, let's join in the comparison value:
select t.*,
(id = min_id) as is_first_occurrence
from t join
(select customer_id, min(id) as min_id
from t
group by customer_id
) tt
on t.customer_id = tt.customer_id;
For most people, a "unique" flag would mean that the overall count is "1", not that this is merely the first appearance. If that is what you want, then you can use similar logic:
select t.*,
(id = min_id) as is_first_occurrence,
(cnt = 1) as is_unique
from t join
(select customer_id, min(id) as min_id, count(*) as cnt
from t
group by customer_id
) tt
on t.customer_id = tt.customer_id;
And, in MySQL 8+, you would use window functions:
select t.*,
(row_number() over (partition by customer_id order by id) = 1) as is_first_occurrence,
(count(*) over (partition by customer_id) = 1) as is_unique
from t;
answered Nov 23 '18 at 12:39
Gordon Linoff
758k35291399
add a comment |
You seem to want to mark the first occurrence as unique, but not others. So, let's join in the comparison value:
select t.*,
(id = min_id) as is_first_occurrence
from t join
(select customer_id, min(id) as min_id
from t
group by customer_id
) tt
on t.customer_id = tt.customer_id;
For most people, a "unique" flag would mean that the overall count is "1", not that this is merely the first appearance. If that is what you want, then you can use similar logic:
select t.*,
(id = min_id) as is_first_occurrence,
(cnt = 1) as is_unique
from t join
(select customer_id, min(id) as min_id, count(*) as cnt
from t
group by customer_id
) tt
on t.customer_id = tt.customer_id;
And, in MySQL 8+, you would use window functions:
select t.*,
(row_number() over (partition by customer_id order by id) = 1) as is_first_occurrence,
(count(*) over (partition by customer_id) = 1) as is_unique
from t;
answered Nov 23 '18 at 12:39
Gordon Linoff
758k35291399
You seem to want to mark the first occurrence as unique, but not others. So, let's join in the comparison value:
select t.*,
(id = min_id) as is_first_occurrence
from t join
(select customer_id, min(id) as min_id
from t
group by customer_id
) tt
on t.customer_id = tt.customer_id;
For most people, a "unique" flag would mean that the overall count is "1", not that this is merely the first appearance. If that is what you want, then you can use similar logic:
select t.*,
(id = min_id) as is_first_occurrence,
(cnt = 1) as is_unique
from t join
(select customer_id, min(id) as min_id, count(*) as cnt
from t
group by customer_id
) tt
on t.customer_id = tt.customer_id;
And, in MySQL 8+, you would use window functions:
select t.*,
(row_number() over (partition by customer_id order by id) = 1) as is_first_occurrence,
(count(*) over (partition by customer_id) = 1) as is_unique
from t;
answered Nov 23 '18 at 12:39
Gordon Linoff
758k35291399
answered Nov 23 '18 at 12:39
Gordon Linoff
758k35291399
answered Nov 23 '18 at 12:39
Gordon Linoff
758k35291399
answered Nov 23 '18 at 12:39
Gordon Linoff
758k35291399
758k35291399
add a comment |
add a comment |
You can try below
select id,a.customerid, case when cnt=1 then 1 else 0 end as Unique_Flag
from tablename a
left join
(select customerid, count(*) as cnt from tablename
group by customerid
)b on a.customerid=b.customerid
answered Nov 23 '18 at 4:48
fa06
11k2917
add a comment |
You can try below
select id,a.customerid, case when cnt=1 then 1 else 0 end as Unique_Flag
from tablename a
left join
(select customerid, count(*) as cnt from tablename
group by customerid
)b on a.customerid=b.customerid
answered Nov 23 '18 at 4:48
fa06
11k2917
add a comment |
You can try below
select id,a.customerid, case when cnt=1 then 1 else 0 end as Unique_Flag
from tablename a
left join
(select customerid, count(*) as cnt from tablename
group by customerid
)b on a.customerid=b.customerid
answered Nov 23 '18 at 4:48
fa06
11k2917
You can try below
select id,a.customerid, case when cnt=1 then 1 else 0 end as Unique_Flag
from tablename a
left join
(select customerid, count(*) as cnt from tablename
group by customerid
)b on a.customerid=b.customerid
answered Nov 23 '18 at 4:48
fa06
11k2917
answered Nov 23 '18 at 4:48
fa06
11k2917
answered Nov 23 '18 at 4:48
fa06
11k2917
answered Nov 23 '18 at 4:48
fa06
11k2917
11k2917
add a comment |
add a comment |
You can use lead function as given below to get the required output.
SELECT ID, CUSTOMER_ID,
CASE
WHEN CUSTOMER_ID != CUSTOMER_ID_NEXT THEN 1
ELSE 0
END AS UNIQUE_FLAG FROM
(SELECT ID, CUSTOMER_ID,LEAD(CUSTOMER_ID, 1, 0) OVER (ORDER BY CUSTOMER_ID) AS CUSTOMER_ID_NEXT FROM TABLE)T
answered Nov 23 '18 at 9:36
Vivek Khandelwal
1246
add a comment |
You can use lead function as given below to get the required output.
SELECT ID, CUSTOMER_ID,
CASE
WHEN CUSTOMER_ID != CUSTOMER_ID_NEXT THEN 1
ELSE 0
END AS UNIQUE_FLAG FROM
(SELECT ID, CUSTOMER_ID,LEAD(CUSTOMER_ID, 1, 0) OVER (ORDER BY CUSTOMER_ID) AS CUSTOMER_ID_NEXT FROM TABLE)T
answered Nov 23 '18 at 9:36
Vivek Khandelwal
1246
add a comment |
You can use lead function as given below to get the required output.
SELECT ID, CUSTOMER_ID,
CASE
WHEN CUSTOMER_ID != CUSTOMER_ID_NEXT THEN 1
ELSE 0
END AS UNIQUE_FLAG FROM
(SELECT ID, CUSTOMER_ID,LEAD(CUSTOMER_ID, 1, 0) OVER (ORDER BY CUSTOMER_ID) AS CUSTOMER_ID_NEXT FROM TABLE)T
answered Nov 23 '18 at 9:36
Vivek Khandelwal
1246
You can use lead function as given below to get the required output.
SELECT ID, CUSTOMER_ID,
CASE
WHEN CUSTOMER_ID != CUSTOMER_ID_NEXT THEN 1
ELSE 0
END AS UNIQUE_FLAG FROM
(SELECT ID, CUSTOMER_ID,LEAD(CUSTOMER_ID, 1, 0) OVER (ORDER BY CUSTOMER_ID) AS CUSTOMER_ID_NEXT FROM TABLE)T
answered Nov 23 '18 at 9:36
Vivek Khandelwal
1246
edited Nov 23 '18 at 13:46
answered Nov 23 '18 at 9:36
Vivek Khandelwal
1246
answered Nov 23 '18 at 9:36
Vivek Khandelwal
1246
answered Nov 23 '18 at 9:36
Vivek Khandelwal
1246
1246
add a comment |
add a comment |
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